SolSec 2.2

# SolSec 2.2 - 2 9 Problems and Solutions Section 2.2(2.11...

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2- 9 Problems and Solutions Section 2.2 (2.11 through 2.25) 2.11 Calculate the constants A and φ for arbitrary initial conditions, x 0 and v 0 , in the case of the forced response given by Equation 2.29. Compare this solution to the transient response obtained in the case of no forcing function (i.e. F 0 = 0). Solution: From equation (2.29) x t Ae t X t xt Ae t A e t X t n nn t d n t dd t d ( ) sin( ) cos( ) ˙( ) sin( ) cos( ) sin( ) =+ + =− + + + −− ζω ωφ ω θ Next apply the initial conditions to these general expressions for position and velocity to get: xA X A X n d ( ) sin cos ˙( ) sin cos sin 0 0 + + φθ ζ φω Solving this system of two equations in two unknowns yields: θω θζω = = tan ( cos ) ( cos ) sin cos sin 1 0 00 0 xX vx X X A d Recall that X has the form X Fm n n = −+ = 0 22 2 2 1 2 2 / () ( ) tan ωω ζω ω and Now if F 0 = 0, then X = 0 and A and φ from above reduce to: = + == ++ tan sin ( ) 1 0 000 2 0 2 2 x A xv x x d n n d d These are identical to the values given in equation (1.38).

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2-10 2.12 Show that Equations 2.20 and 2.21 are equivalent by verifying Equations 2.21 and 2.22. Solution: From equation (2.20) and expanding the trig relation yields xX t X t t Xt X t p AB =− = + [] =+ cos( ) cos cos sin sin ( cos )cos ( sin )sin ωθ ω θ θω 12 43 41 2 4 Now with A and B defined as indicated, the magnitude is computed: XA B 22 and B A X X B A =⇒ = sin cos tan θ 1
2-11 2.13 Plot the solution of Equation 2.19 for the case that m = 1 kg, ζ = 0.01, ω n = 2 rad/s. F 0 = 3 N, and = 10 rad/s, with initial conditions x 0 = 1 m and v 0 = 1 m/s. Solution: The particular solution is given in equations (2.28) and (2.29). Substitution of the values given yields: xt p =+ 0 03125 10 0 004176 . cos( . ) . Then the total solution has the form: Ae t t t ( ) sin( ) . cos( . ) . + + 02 2 0 03125 10 0 004176 φ Differentiating then yields ˙( ) . sin( ) cos( ) . sin( . ) .. x t Ae t Ae t t tt =− + + + + 0 2 2 2 2 0 3125 10 0 004176 φφ Apply the initial conditions to get: xA A ( ) sin . ˙( ) . sin cos . 0 1 0 03125 0 1 0 2 2 0 3125 == + + Solving yields: A = 1.095 m and =1.086 rad. So the solution and plot become (using Mathcad)

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2-12 2.14 A 100 kg mass is suspended by a spring of stiffness 30 × 10 3 N/m with a viscous damping constant of 1000 Ns/m. The mass is initially at rest and in equilibrium. Calculate the steady-state displacement amplitude and phase if the mass is excited by a harmonic force of 80 N at 3 Hz. Solution: Given m = 100kg, k =30,000 N/m, c = 1000 Ns/m, F 0 = 80 N and ω = 6 π rad/s: f F m k m c km X n 0 0 22 2 2 80 100 0 8 17 32 2 0 289 08 17 32 36 2 0 289 17 32 6 0 0 0041 == = = = = + () + = ., . . . .( . ) ( . ) ( . m/s rad/s m 2 ω ζ ππ Next compute the angle from φ = tan . . 1 188 702 55 323 Since the denominator is negative the angle must be found in the 4 th quadrant. To find this use Window 2.3 and then in Matlab type atan2(188.702,-55.323) or use the principle value and add π to it. Either way the phase is φ =1.856 rad.
2-13 2.15 Plot the total solution of the system of Problem 2.14 including the transient.

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SolSec 2.2 - 2 9 Problems and Solutions Section 2.2(2.11...

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